已知椭圆的方程为x²/9+y²=1,过左焦点作倾斜角为π/6的直线,交椭圆于A,B两点,求AB长.

椭圆左焦点为F1(2√2,0),焦点弦斜率为k=tan(π/6)=1/√3
则焦点弦方程为 y=(x-2√2)/√3 或 x=√3y+2√2
代入椭圆可得 (√3y+2√2)^2+9y^2=9
整理得 12y^2+4√6y-1=0
设两交点为A(x1,y1),B(x2,y2),则由韦达定理可得
y1+y2=-√6/3,y1y2=-1/12; x1+x2=√3(y1+y2)+4√2=3√2
x1x2=(√3y1+2√2)(√3y2+2√2)=3y1y2+2√6(y1+y2)+8=-1/4-4+8=15/4
∴AB=√[(x1-x2)^2+(y1-y2)^2]
=√[(x1+x2)^2-4x1x2+(y1+y2)^2-4y1y2]
=√[(3√2)^2-4*(15/4)+(-√6/3)^2-4*(-1/12)]
=√(3+1)
=2